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VICSpecialist Mathematics2024Exam 1

VCE Specialist Mathematics 2024 Exam 1

Worked solutions to every question in the 2024 VCE Specialist Mathematics Examination 1 (40 marks, no calculator), checked against the VCAA external assessment report, with the common errors the report flagged.

Marks: 40
Time: 60 min
Authority: VCAA
Updated: 2026-09-23

Open the official PDF Official marking notes

Every question from the 2024 VCE Specialist Mathematics Examination 1, the technology-free paper, with a full worked solution. Solutions sit behind a Show worked solution toggle so you can attempt each question first. For the calculator paper, see the 2024 Examination 2 walkthrough.

How to use this page

Questions are from the 2024 VCE Specialist Mathematics Examination 1, copyright Victorian Curriculum and Assessment Authority (VCAA). Each question is summarised in a line here; open the official examination PDF for the exact wording and diagrams.
Solutions are original ExamExplained working, checked line by line and compared with the 2024 Specialist Mathematics Examination 1 external assessment report (Word document). Both files are listed on the VCAA Specialist Mathematics examinations page.
The From the report note under each solution summarises what the assessors said about that question.

Structure and timing

Examination 1 is 40 marks in 60 minutes (plus 15 minutes reading time), with no calculator or notes; a formula sheet is provided. There were 10 questions in 2024. According to the report, students did well on the direct proof, vectors, volumes of revolution, separable differential equations and implicit differentiation, and struggled with graph sketching (Q3c), kinematics (Q9b) and the distance between skew lines (Q10).

Questions and worked solutions

Question 1 (4 marks)

Let $f(z) = 3z^3 + 2iz^2 + 3z + 2i$ , $z \in C$ .

a
Verify that $3z + 2i$ is a factor of $f(z)$ . (1 mark)
b
Hence, or otherwise, solve $f(z) = 0$ , giving answers in Cartesian form. (2 marks)
c
Plot the solutions on an Argand diagram. (1 mark)

Show worked solution

a. [1 mark]. Group the terms: $f(z) = z^2(3z + 2i) + (3z + 2i) = (3z + 2i)(z^2 + 1)$ . (Equivalently, $f\!\left(-\tfrac{2i}{3}\right) = 0$ by the factor theorem.)

b. [2 marks]. $(3z + 2i)(z^2 + 1) = 0$ gives $z = -\tfrac23i$ or $z^2 = -1$ , so

z = -\frac23i, \quad z = i, \quad z = -i.

c. [1 mark]. All three solutions lie on the imaginary axis.

From the report. In part b some wrote down solutions without showing an equation being solved, and some wrongly applied the conjugate root theorem (the coefficients are not all real). In part c some plotted the points on the real axis.

Question 2 (3 marks)

Prove, using a direct proof, that if $x$ is an odd integer then $2x^2 - 3x - 7$ is even. (3 marks)

Show worked solution

[3 marks]. Let $x$ be odd, so $x = 2k + 1$ for some integer $k$ . Then

2x^2 - 3x - 7 = 2(4k^2 + 4k + 1) - 3(2k + 1) - 7 = 8k^2 + 2k - 8 = 2(4k^2 + k - 4).

Since $4k^2 + k - 4$ is an integer, $2x^2 - 3x - 7$ is a multiple of 2, so it is even.

From the report. Well answered (65% full marks); occasional algebra slips.

Question 3 (6 marks)

Let $f(x) = \dfrac{(x - 1)^2}{(x + 1)^2}$ , $x \ne -1$ , which can be written as $A + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$ with $A, B, C \in Z$ .

a
Show that $A = 1$ , $B = -4$ and $C = 4$ . (1 mark)
b
Find the coordinates of the one turning point. (2 marks)
c
Sketch $y = f(x)$ , labelling asymptotes and axial intercepts. (3 marks)

Show worked solution

a. [1 mark]. Write the numerator in powers of $(x + 1)$ : $(x - 1)^2 = \big((x + 1) - 2\big)^2 = (x + 1)^2 - 4(x + 1) + 4$ . Dividing by $(x + 1)^2$ :

f(x) = 1 - \frac{4}{x + 1} + \frac{4}{(x + 1)^2}.

b. [2 marks].

f'(x) = \frac{4}{(x + 1)^2} - \frac{8}{(x + 1)^3} = \frac{4(x - 1)}{(x + 1)^3} = 0 \implies x = 1.

$f(1) = 0$ , so the turning point is $(1, 0)$ .

c. [3 marks]. Asymptotes $x = -1$ and $y = 1$ . The graph touches the $x$ -axis at $(1, 0)$ and crosses the $y$ -axis at $(0, 1)$ . Since $f(x) - 1 = \tfrac{-4x}{(x + 1)^2}$ , the graph is above $y = 1$ for $x < 0$ (so the whole left branch lies above the asymptote) and below it for $x > 0$ .

From the report. Part c was done poorly (10% full marks): extra or wrong asymptotes, unlabelled features, and missing left branches were common. Evaluating a few points on the left helped students get its shape right.

Question 4 (4 marks)

$\mathbf{a} = 3\mathbf{j} + 3\mathbf{k}$ , $\mathbf{b} = 2\mathbf{i} - \mathbf{j} - 2\mathbf{k}$ and $\mathbf{c} = n\mathbf{i} + 2\mathbf{j} + \mathbf{k}$ , $n \in Z$ .

a. Find the angle between $\mathbf{a}$ and $\mathbf{b}$ . (2 marks)

b. Find all $n$ such that $\mathbf{a}\cdot\mathbf{c} = |\mathbf{a}\times\mathbf{c}|$ . (2 marks)

Show worked solution

a. [2 marks]. $\mathbf{a}\cdot\mathbf{b} = 0 - 3 - 6 = -9$ , $|\mathbf a| = 3\sqrt2$ and $|\mathbf b| = 3$ :

\cos\theta = \frac{-9}{9\sqrt2} = -\frac{1}{\sqrt2} \implies \theta = \frac{3\pi}{4} \ (135^\circ).

b. [2 marks]. $\mathbf{a}\cdot\mathbf{c} = 6 + 3 = 9$ and

\mathbf{a}\times\mathbf{c} = \begin{vmatrix}\mathbf i & \mathbf j & \mathbf k \\ 0 & 3 & 3 \\ n & 2 & 1\end{vmatrix} = -3\mathbf{i} + 3n\mathbf{j} - 3n\mathbf{k}, \qquad |\mathbf{a}\times\mathbf{c}| = \sqrt{9 + 18n^2}.

$\sqrt{9 + 18n^2} = 9$ gives $n^2 = 4$ , so $n = -2$ or $n = 2$ .

From the report. In part a many found $\cos\theta = -\tfrac{1}{\sqrt2}$ but then gave the wrong angle, often $\tfrac{\pi}{4}$ . In part b some gave only one value of $n$ .

Question 5 (3 marks)

The curve $y = \sqrt{k - \dfrac{1}{x^2}}$ , $1 \le x \le \dfrac{k}{2}$ , $k > 2$ , is rotated about the $x$ -axis to give volume $\dfrac{7\pi}{2}$ . Show that $k^3 - 2k^2 - 9k + 4 = 0$ . (3 marks)

Show worked solution

[3 marks].

V = \pi\int_1^{k/2}\left(k - \frac{1}{x^2}\right)dx = \pi\left[kx + \frac1x\right]_1^{k/2} = \pi\left(\frac{k^2}{2} + \frac2k - k - 1\right).

Setting $V = \tfrac{7\pi}{2}$ and multiplying by $2k$ :

k^3 + 4 - 2k^2 - 2k = 7k \implies k^3 - 2k^2 - 9k + 4 = 0.

From the report. Mostly well done. Some used a surface area formula instead, and sign errors when subtracting the lower-terminal value (ending with $-k + 1$ instead of $-k - 1$ ) were common.

Question 6 (5 marks)

Stages 1, 2 and 3 of making a weed trimmer take independent normal times $W_1, W_2, W_3$ hours with means 1.0, 1.5, 2.0, standard deviations 0.3, 0.4, 0.5, and costs $10, $20 and $15 per hour.

a
Find the mean and variance of the total time. (1 mark)
b
Find the variance of the total cost. (2 marks)
c
Find the probability that Stage 2 takes less time than Stage 1, to two decimal places, using $\Pr(-1 < Z < 1) = 0.68$ . (2 marks)

Show worked solution

a. [1 mark]. $T = W_1 + W_2 + W_3$ : mean $1 + 1.5 + 2 = 4.5$ h, variance $0.3^2 + 0.4^2 + 0.5^2 = 0.5$ h².

b. [2 marks]. $C = 10W_1 + 20W_2 + 15W_3$ , so

\text{var}(C) = 10^2(0.09) + 20^2(0.16) + 15^2(0.25) = 9 + 64 + 56.25 = 129.25.

c. [2 marks]. Let $D = W_2 - W_1$ : mean $0.5$ , variance $0.16 + 0.09 = 0.25$ , standard deviation $0.5$ . Then

\Pr(D < 0) = \Pr(Z < -1) = \frac{1 - 0.68}{2} = 0.16.

From the report. In part b many forgot to square the costs or the standard deviations. In part c some slipped with the mean or variance of the difference, or gave the answer without working.

Question 7 (4 marks)

Solve $x + 2y\sqrt{x^2 + 1}\,\dfrac{dy}{dx} = 0$ with $y(0) = -2$ , giving $y$ as a function of $x$ . (4 marks)

Show worked solution

[4 marks]. Separate the variables:

\int 2y\,dy = -\int\frac{x}{\sqrt{x^2 + 1}}\,dx \implies y^2 = -\sqrt{x^2 + 1} + c.

$y(0) = -2$ gives $4 = -1 + c$ , so $c = 5$ and $y^2 = 5 - \sqrt{x^2 + 1}$ . Since $y(0) < 0$ , take the negative root:

y = -\sqrt{5 - \sqrt{x^2 + 1}}.

From the report. Most recognised a separable equation. Marks were lost for leaving the answer as $y^2 = \ldots$ instead of $y$ as a function of $x$ , and for choosing the positive square root.

Question 8 (4 marks)

Consider $x^2y^2 + xy = 2$ .

a. Use implicit differentiation to show that $\dfrac{dy}{dx} = -\dfrac{y}{x}$ , given $2xy \ne -1$ . (2 marks)

b. Find all points on the graph where the tangent has slope $-1$ . (2 marks)

Show worked solution

a. [2 marks].

2xy^2 + 2x^2y\frac{dy}{dx} + y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx}\,x(2xy + 1) = -y(2xy + 1).

Since $2xy + 1 \ne 0$ , divide it out: $\dfrac{dy}{dx} = -\dfrac{y}{x}$ .

b. [2 marks]. $-\tfrac{y}{x} = -1$ means $y = x$ . Substituting into the relation: $x^4 + x^2 - 2 = 0$ , so $(x^2 + 2)(x^2 - 1) = 0$ and $x = \pm1$ . The points are $(1, 1)$ and $(-1, -1)$ .

From the report. In part a the factorisation of $2xy + 1$ had to be shown. In part b some found $y = x$ but did not substitute back into the relation, or added incorrect points.

Question 9 (4 marks)

A car's velocity $v$ km/h and position $x$ km satisfy $v^2 = 1600 + \dfrac{672}{\pi}\arccos\!\left(\dfrac{x}{20}\right)$ for $-15 \le x \le 15$ , $v \ge 0$ . A device at $x = 0$ is activated if the speed is 10% or more above the 40 km/h limit.

a. Determine, with evidence, whether the device is activated. (1 mark)

b. Find the acceleration when $x = 12$ , in the form $\tfrac{k}{\pi}$ , $k \in Z$ . (3 marks)

Show worked solution

a. [1 mark]. The device needs $v \ge 44$ km/h. At $x = 0$ , $\arccos(0) = \tfrac{\pi}{2}$ , so $v^2 = 1600 + 336 = 1936 = 44^2$ and $v = 44$ . The device is activated.

b. [3 marks]. Use $a = \dfrac{d}{dx}\left(\tfrac12v^2\right)$ :

a = \frac12 \cdot \frac{672}{\pi} \cdot \frac{-1}{\sqrt{1 - \frac{x^2}{400}}} \cdot \frac{1}{20} = -\frac{336}{20\pi\sqrt{1 - \frac{x^2}{400}}}.

At $x = 12$ , $\sqrt{1 - \tfrac{144}{400}} = \tfrac45$ , so $a = -\dfrac{336}{20\pi} \times \dfrac54 = -\dfrac{21}{\pi}$ km/h². So $k = -21$ .

From the report. In part a many simply assumed any speed over 40 km/h triggers the device; the working had to show $v^2 = 1936 = 44^2$ at $x = 0$ . In part b the factor $\tfrac12$ or the negative sign from the derivative of $\arccos$ was often lost.

Question 10 (3 marks)

$l_1: \mathbf{r}_1(\lambda) = \mathbf{i} + m\mathbf{k} + \lambda(\mathbf{i} + 2\mathbf{j} + \mathbf{k})$ and $l_2: \mathbf{r}_2(\mu) = 2\mathbf{i} - \mathbf{k} + \mu(-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})$ , $m \ne -\tfrac45$ . If the shortest distance between the skew lines is $\dfrac{14}{\sqrt{35}}$ , find $m$ . (3 marks)

Show worked solution

[3 marks]. A vector perpendicular to both lines is

\mathbf{n} = (\mathbf{i} + 2\mathbf{j} + \mathbf{k}) \times (-\mathbf{i} + 3\mathbf{j} + 2\mathbf{k}) = \mathbf{i} - 3\mathbf{j} + 5\mathbf{k}, \qquad |\mathbf n| = \sqrt{35}.

The shortest distance is the projection onto $\mathbf n$ of the vector joining a point on each line. From $(1, 0, m)$ to $(2, 0, -1)$ the vector is $\mathbf{i} + (-1 - m)\mathbf{k}$ :

d = \frac{\left|1 - 5(1 + m)\right|}{\sqrt{35}} = \frac{|-4 - 5m|}{\sqrt{35}} = \frac{14}{\sqrt{35}} \implies |5m + 4| = 14.

So $m = 2$ or $m = -\dfrac{18}{5}$ .

From the report. Poorly answered (59% scored zero). Some tried to recall a formula from memory; a quick diagram of skew lines and the common perpendicular helps. Many took only one sign and so found only one value of $m$ .

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Structure and timing

Questions and worked solutions

Question 1 (4 marks)

Question 2 (3 marks)

Question 3 (6 marks)

Question 4 (4 marks)

Question 5 (3 marks)

Question 6 (5 marks)

Question 7 (4 marks)

Question 8 (4 marks)

Question 9 (4 marks)

Question 10 (3 marks)

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